I was wondering if someone could check the problems for me? Thanks I can go back and show work if needed..this was just easier to type out.
1)An object has a mass of 615 grams and a volume of 105cm3.What is the density of the object?
answer 5.86 gm/cm3
2) A player uses a hockey stick to increase the speed of a 0.200 kg hockey puck by 6 m/s in 2 seconds.?
a) how much did the hockey puck accelerate?
answer-3 m/s2
b) how much force was exerted on the puck?
answer .6N
c) how much force did the puck exert on the hockey stick?
answer -.6 N
3) Find the net force produced by a 40 newton force and a 50 newton in each case
A)both forces act in the same direction
B)bother forces act in opposite directions
I am not sure :-/
4)A car of mass 900kg moves at 10 m/s. What is the breaking force required to stop the car in 5 seconds
Not sure either
5)A 10 kg go-cart is moving at a speed of 3 m/s. what is the kinetic energy?
45J
6) if an engine must do 500 joules of work to pull a 10 kg load up to a height of 4 meters, what is the efficiency of the engine?
80%
Thanks
oh two more I forgot sorry
1)Suppose you want to heat 40 kg of water by 20c how many joules of heat are required?
3344 J
2)How many joules of heat are required to melt 40kg of ice at 0 degrees C?
13360 J, akk I am thinking that is off
The density is ok but g is grams and not gm.
On supolementary 1, I think you forgot to multiply 40 kg by 1000 to change to grams.
40 kg x 1000 g/kg x 4.184 J/g*C x 20 degrees C = ??
2. 40 kg x 1000 g/kg x heat of fusion = ??
You lost me...
1)Suppose you want to heat 40 kg of water by 20c how many joules of heat are required?
3344 J
40,000 g x 4.184 J/g*C x 20 C = joules heat required to heat 40 kg water by 20 degrees C.
2)How many joules of heat are required to melt 40kg of ice at 0 degrees C?
13360 J, akk I am thinking that is off
J = 40,000 g x heat of fusion ice = 40,000 g x 334 J/g = ??
Look up the heat of fusion of water. I may not have remembered 334 J/g correctly but I think that is right.
1) An object has a mass of 615 grams and a volume of 105 cm3. To find the density of the object, divide the mass by the volume:
Density = Mass / Volume
Density = 615 grams / 105 cm3
Density = 5.86 grams/cm3
So, the density of the object is 5.86 grams/cm3.
2a) To find the acceleration of the hockey puck, use the equation:
Acceleration = Change in velocity / Time
Acceleration = 6 m/s / 2 s
Acceleration = 3 m/s2
So, the hockey puck accelerated at a rate of 3 m/s2.
2b) To find the force exerted on the puck, use Newton's second law of motion:
Force = Mass x Acceleration
Force = 0.200 kg x 3 m/s2
Force = 0.6 N
So, a force of 0.6 N was exerted on the puck.
2c) By Newton's third law of motion, the force exerted by the puck on the hockey stick is equal in magnitude but opposite in direction to the force exerted on the puck. So, the force exerted by the puck on the hockey stick is -0.6 N.
3a) When both forces act in the same direction, the net force is the sum of the forces. So, the net force would be 40 N + 50 N = 90 N.
3b) When both forces act in opposite directions, the net force is the difference between the forces. So, the net force would be 50 N - 40 N = 10 N.
4) To find the braking force required to stop the car, you can use Newton's second law of motion:
Force = Mass x Acceleration
Given:
Mass = 900 kg
Initial velocity = 10 m/s
Final velocity = 0 m/s
Time = 5 seconds
The acceleration can be calculated using the equation:
Acceleration = (Final velocity - Initial velocity) / Time
Acceleration = (0 m/s - 10 m/s) / 5s
Acceleration = -2 m/s2 (negative sign indicates deceleration)
Now, use Newton's second law to calculate the braking force:
Force = Mass x Acceleration
Force = 900 kg x -2 m/s2
Force = -1800 N
So, the braking force required to stop the car is -1800 N (negative sign indicates opposite direction to the motion).
5) To find the kinetic energy of the go-cart, use the equation:
Kinetic Energy = (1/2) x Mass x Velocity^2
Given:
Mass = 10 kg
Velocity = 3 m/s
Kinetic Energy = (1/2) x 10 kg x (3 m/s)^2
Kinetic Energy = 45 J
So, the kinetic energy of the go-cart is 45 J.
6) To find the efficiency of the engine, you can use the equation:
Efficiency = (Useful Work Output / Total Work Input) x 100%
Given:
Useful Work Output = 500 J
Load = 10 kg
Height = 4 meters
Total Work Input = Force x Distance
Total Work Input = (Mass x Gravity) x Distance
Total Work Input = (10 kg x 9.8 m/s2) x 4 m
Total Work Input = 392 J
Now, plug in the values to calculate the efficiency:
Efficiency = (500 J / 392 J) x 100%
Efficiency = 128% (rounded to nearest whole number)
So, the efficiency of the engine is 128%.
Regarding the supplementary questions:
1) To heat 40 kg of water by 20°C, you need to multiply the mass of water by the specific heat capacity of water and the change in temperature:
Heat = Mass x Specific Heat Capacity x Change in Temperature
Heat = 40 kg x 4.184 J/g°C x 20°C
Heat = 33,440 J (or 33.44 kJ)
So, 33,440 J (or 33.44 kJ) of heat is required.
2) To melt 40 kg of ice at 0°C, you need to multiply the mass of ice by the heat of fusion of ice:
Heat = Mass x Heat of Fusion
Heat = 40 kg x 334 J/g
Heat = 13,360 J (or 13.36 kJ)
So, 13,360 J (or 13.36 kJ) of heat is required.
1) To find the density of an object, you need to divide its mass by its volume. In this case, the mass is given as 615 grams and the volume is given as 105 cm^3. So, the density can be calculated as:
Density = Mass / Volume
Density = 615 g / 105 cm^3
Density = 5.857 g/cm^3
So, the density of the object is 5.857 g/cm^3.
2a) The acceleration of the hockey puck can be found using the formula:
Acceleration = Change in velocity / Time taken
Acceleration = 6 m/s / 2 s
Acceleration = 3 m/s^2
So, the hockey puck accelerated at 3 m/s^2.
2b) The force exerted on the puck can be found using Newton's second law of motion:
Force = Mass x Acceleration
Force = 0.200 kg x 3 m/s^2
Force = 0.6 N
So, the force exerted on the puck is 0.6 N.
2c) According to Newton's third law of motion, the force exerted by the puck on the hockey stick will be equal in magnitude but opposite in direction to the force exerted by the hockey stick on the puck. So, the force exerted by the puck on the hockey stick will also be -0.6 N.
3) The net force produced by multiple forces acting on an object can be found by adding or subtracting the forces depending on their directions.
A) If both forces act in the same direction, simply add their magnitudes to find the net force. So, the net force would be 40 N + 50 N = 90 N.
B) If both forces act in opposite directions, subtract the magnitude of the smaller force from the magnitude of the larger force. So, the net force would be 50 N - 40 N = 10 N.
4) To find the braking force required to stop a car, you can use Newton's second law of motion:
Force = Mass x Acceleration
Acceleration = Change in velocity / Time taken
In this case, the velocity needs to be reduced from 10 m/s to 0 m/s in 5 seconds. So, the change in velocity is 10 m/s.
Acceleration = Change in velocity / Time taken
Acceleration = 10 m/s / 5 s
Acceleration = 2 m/s^2
Now, we can find the braking force:
Force = Mass x Acceleration
Force = 900 kg x 2 m/s^2
Force = 1800 N
So, the braking force required to stop the car is 1800 N.
5) The kinetic energy of an object can be found using the formula:
Kinetic Energy = (1/2) x Mass x Velocity^2
In this case, the mass of the go-cart is given as 10 kg and the speed is given as 3 m/s. Plugging in these values into the formula, we get:
Kinetic Energy = (1/2) x 10 kg x (3 m/s)^2
Kinetic Energy = 45 J
So, the kinetic energy of the go-cart is 45 J.
6) The efficiency of an engine can be found using the formula:
Efficiency = (Useful work output / Total work input) x 100
In this case, the useful work done by the engine is given as 500 J and the total work input is the work required to lift the load, which can be calculated using the formula:
Work = Force x Distance
The force is equal to the weight of the load, which is given by the formula:
Weight = Mass x Gravity
The mass of the load is 10 kg and the height is 4 meters. Plugging in these values, we get:
Work = (10 kg x Gravity) x 4 m
Now, we can calculate the total work input and the efficiency:
Total work input = Work = (10 kg x Gravity) x 4 m
Efficiency = (Useful work output / Total work input) x 100
Efficiency = (500 J / ((10 kg x Gravity) x 4 m)) x 100
Now, you just need to substitute the value of the acceleration due to gravity (which is approximately 9.8 m/s^2) and calculate the efficiency.